Problems on Trains Questions

Multiple choice
  1. 136.6 seconds

  2. 129.6 seconds

  3. 99.6 seconds

  4. 106 seconds

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Opposite direction: relative speed = 70 + 56 = 126 km/h = 35 m/s. Distance = (L + L/3) = 4L/3. Time = 14.4s, so 4L/3 = 35 × 14.4, giving L = 378 m. Same direction: relative speed = 70 - 56 = 14 km/h = 35/9 m/s. Time = 4L/3 ÷ 35/9 = 129.6 seconds. This is a standard relative speed application.

Multiple choice
  1. $600 m, 72 km/h$
  2. $550 m, 75 km/h$
  3. $500 m, 70 km/h$
  4. $650 m, 74 km/h$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let L be train length, S be speed. For 400m platform: (L+400)/50 = S. For 600m platform: (L+600)/60 = S. Equating: (L+400)/50 = (L+600)/60. Cross-multiply: 60(L+400) = 50(L+600). 60L + 24000 = 50L + 30000. 10L = 6000, L = 600m. Speed = (600+400)/50 = 1000/50 = 20 m/s = 20×(18/5) = 72 km/h. Option A is correct. Option B(550m, 75 km/h), C(500m, 70 km/h), and D(650m, 74 km/h) don't satisfy the equations.

Multiple choice
  1. 200 m 200 मी

  2. 300 m 300 मी

  3. 400 m 400 मी

  4. 500 m 500 मी

  5. None of these इनमें से कोई नहीं

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

First, find the train's length. Speed = 72 km/h = 72 × 1000/3600 = 20 m/s. Length of train = speed × time to cross pole = 20 × 10 = 200 m. When crossing the platform, total distance = train length + platform length = 20 × 25 = 500 m. Therefore, platform length = 500 - 200 = 300 m.

Multiple choice
  1. 100 m. 100 मी.

  2. 150 m. 150 मी.

  3. 180 m. 180 मी.

  4. 200 m. 200 मी.

  5. 240 m. 240 मी.

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let train length = L meters. Speed = L/20 m/s (from pole crossing). For platform: (L+100)/speed = 30 seconds. Substituting: (L+100)/(L/20) = 30 → 20(L+100)/L = 30 → 20L + 2000 = 30L → 10L = 2000 → L = 200m.

Multiple choice
  1. 2 m/s

  2. 4 m/s

  3. 6 m/s

  4. 10 m/s

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let faster train speed = v1 m/s, slower train speed = v2 m/s. Same direction: (v1 - v2) = (90+90)/18 = 180/18 = 10 m/s. Opposite direction: (v1 + v2) = (90+90)/9 = 180/9 = 20 m/s. Adding these equations: 2v1 = 30, so v1 = 15 m/s. Then v2 = 20 - 15 = 5 m/s. Difference = v1 - v2 = 15 - 5 = 10 m/s. The answer is 10 m/s.

Multiple choice
  1. 250 m.\मी.

  2. 150 m.\मी.

  3. 300 m.\मी.

  4. 200 m.\मी.

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Relative speed (same direction) = 50 - 40 = 10 km/hr = 25/9 m/s. Time = 2 min 24 sec = 144 sec. Total distance = (25/9) × 144 = 400 m. Length of second train = 400 - 150 = 250 m. Option A is correct.

Multiple choice
  1. 2 : 1

  2. 2 : 3

  3. 3 : 2

  4. 4 : 3

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let train speed = t km/h and car speed = c km/h. Case 1: 120/t + 1680/c = 8. Case 2: 200/t + 1600/c = 8 hours 20 minutes = 25/3 hours. Solving these equations: t = 50 km/h, c = 25 km/h. Ratio t:c = 50:25 = 2:1.

Multiple choice
  1. Quantity I > Quantity II

  2. Quantity I ≥ Quantity II

  3. Quantity I > Quantity I

  4. Quantity II ≥ Quantity I

  5. Quantity I = Quantity II or Relation cannot be established

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Quantity I: Train's relative speed to man = (Train speed - 6 kmph). Distance = 175 m, Time = 10 sec. Relative speed in m/s = 175/10 = 17.5 m/s = 63 kmph. Train speed = 63 + 6 = 69 kmph. Quantity II: Average speed = 2 × 65 × 70 / (65 + 70) = 9100 / 135 ≈ 67.41 kmph. Since 69 > 67.41, Quantity I > Quantity II, so Option A is correct.

Multiple choice
  1. 125 : 126

  2. 125 : 9

  3. 9 : 125

  4. 126 : 125

  5. None of these

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Statement 1: Train 2 takes 18s to cross man. Train 1 takes 50% less = 9s. Speed ratio 4:5, let v1=4x, v2=5x. Lengths: L1=4x×9=36x, L2=5x×18=90x. Same direction crossing time = (36x+90x)/(5x-4x) = 126x/x = 126s. Statement 2: First train length L, speed 54km/h=15m/s. Crosses L/3 platform in 60s: L+L/3 = 15×60, 4L/3 = 900, L=675m. Second train length = 2×L/3 = 450m. Speed of first = 15m/s, second = (3/5)×15 = 9m/s. Same direction crossing time = (675+450)/(15-9) = 1125/6 = 187.5s. Ratio = 126:187.5 = 126:125. ✓

Multiple choice
  1. Only Statement A

  2. Statement B and Statement C together

  3. Statement A and Statement C together

  4. All statements are required

  5. Only Statement B

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To find arrival time, need travel time and departure time. Statement B gives departure (11:15 am) and distance (567 km). Statement C gives train length and pole-crossing time, allowing speed calculation (19.5 m/s). Speed × distance gives travel time. Statement A alone is insufficient.

Multiple choice
  1. 4 m/s./मी./से.

  2. 8 m/s./मी./से.

  3. 12 m/s./मी./से.

  4. 14 m/s./मी./से.

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let speeds be s (slower) and f (faster). Same direction: (f-s) × 20 = 240 (sum of lengths). Opposite direction: (f+s) × 12 = 240. From opposite direction: f+s = 20. From same direction: f-s = 12. Adding: 2f = 32, so f = 16 m/s. Then s = 20 - 16 = 4 m/s. Relative speed changes based on direction - subtract when same direction, add when opposite.

Multiple choice
  1. 54 km/hr. /किमी./घंटा

  2. 60 km/hr. /किमी./घंटा

  3. 36 km/hr. /किमी./घंटा

  4. 48 km/hr. /किमी./घंटा

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For opposite direction: (v1+v2) = 1500/60 = 25 m/s (sum of speeds). For same direction: |v1-v2| = 1500/300 = 5 m/s (difference in speeds). Solving: v1+v2 = 25 and v1-v2 = 5 (assuming v1 > v2). Adding: 2v1 = 30, so v1 = 15 m/s = 54 km/hr. The faster train travels at 54 km/hr.

Multiple choice
  1. 66 km./किमी.

  2. 72 km./किमी.

  3. 78 km./किमी.

  4. 81 km./किमी.

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let train length = L and speed = v km/hr. For person 1 (4.5 km/hr): L = 8.4 × (v - 4.5) × 5/18. For person 2 (5.4 km/hr): L = 8.5 × (v - 5.4) × 5/18. Equating: 8.4(v - 4.5) = 8.5(v - 5.4). Solving: 8.4v - 37.8 = 8.5v - 45.9, giving v = 81 km/hr. Converting speeds to m/s and back is essential - the small time difference (0.1s) determines the speed.

Multiple choice
  1. If the data in statement I alone is sufficient to answer the question.

  2. If the data in statement II alone is sufficient to answer the question.

  3. If the data either in statement I alone or statement II alone are sufficient to answer the question.

  4. If the data given in both I and II together are not sufficient to answer the question.

  5. If the data in both the statements I and II together are necessary to answer the question.

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Statement I: Train crosses 400m train in 25 seconds moving opposite. Relative speed = (L + 400)/25, but we don't know either train's speed. Statement II: Train crosses 750m bridge in 45 seconds. Speed = (L + 750)/45. Even together, we have two unknowns (our train's length L and speed v) but the relationship involves another train's unknown speed. Statement I doesn't give us the other train's speed, so even combining both statements is insufficient to find L.

Multiple choice
  1. Quantity I > Quantity II

  2. Quantity I ≥ Quantity II

  3. Quantity II > Quantity I

  4. Quantity II ≥ Quantity I

  5. Quantity I = Quantity II or Relation cannot be established

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Quantity I: Upstream speed = 4km/(0.5hr) = 8km/h, downstream = 4km/(1/3hr) = 12km/h. Boat speed = (8+12)/2 = 10km/h, current = 2km/h. If current doubles to 4km/h, new downstream = 14km/h, time = 4/14hr = 17.14min. Quantity II: Total distance = 200+600 = 800m in 20s, so speed = 40m/s. For 480m platform, distance = 200+480 = 680m, time = 680/40 = 17s. Since 17.14 > 17, Quantity I > Quantity II.